Basis for the range of the matrix
By Sophia Bowman
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I did the first part properly and showed that the rank is 2, but putting this matrix into a reduced row echleon form. For the second part, I get the wrong basis vectors simply because of something weird. Look at it :
^^ Look at this picture, there is in the reduced echleon form numbers on the right side which are : 1,-15, -17,-6 . Where did these values come from O.o the way I did was (I started like) :
$$\left[\begin{array}{cccc|c} 1 & 1 & 5 & 7 & 0\\ 3 & 9 & 17 & 25 & 0 \\ 1 & 7 & 7 & 11&0\\3&6&16&23&0\end{array}\right]$$
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$\begingroup$A way or obtaining the basis is reducing the transformation matrix and look at the pivot columns. In this case, the pivot numbers are $m1,m2$. Therefore the basis of the transformation are columns $m1,m2$ of $M$, because $m3,m4$ are linear combinations of $m1,m2$.
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